Subobject Lattices Generators and the Grothendieck Axioms

Let $j:J\rightarrowtail A$ be the image inclusion of $[b,c]$. Since $[b,c]{\iota }_B=b$ and $[b,c]{\iota }_C=c$ for the biproduct injections, the factorization of $[b,c]$ through $j$ makes both $b$ and $c$ factor through $j$. Thus $B\le J$ and $C\le J$, so $J$ is an upper bound of the two subobjects.

Let $n:N\rightarrowtail A$ be any common upper bound. Then $b=nu$ and $c=nv$ for suitable $u:B\to N$ and $v:C\to N$. By the universal property of $B\oplus C$, the induced map satisfies $[b,c]=n[u,v]$. Now [L2] says that the image inclusion $j$ factors through every monomorphism through which $[b,c]$ factors, so $J\le N$.

Step 1.1 gives that $J$ is an upper bound, and step 1.2 gives that it lies below every upper bound. By [L3], this is exactly the least-upper-bound claim.

Form a pullback square tikzcd P \arrow[r, "q"] \arrow[d, "p"'] & C \arrow[d, "c"] \\ B \arrow[r, "b"'] & A. By [L2], the pullback leg $p$ is monic, so the composite $m:=bp=cq:P\to A$ is monic as well. Because $m$ factors through both $b$ and $c$, it is a lower bound of the two subobjects.

Let $n:N\rightarrowtail A$ be any lower bound. Then $n=bu=cv$ for some $u:N\to B$ and $v:N\to C$. By the pullback universal property [L1], there is a unique $w:N\to P$ with $pw=u$ and $qw=v$. Therefore $mw=bu=n$, so $n$ factors through $m$. Hence $n\le m$.

Steps 1.1 and 1.2 show that $m$ is a lower bound above every other lower bound. By [L3], the pullback subobject is exactly $B\wedge C$.

By [L1] and [L2], every pair of subobjects of $A$ has a join and a meet. That gives the binary lattice operations.

By [L4], the unique map $0\to A$ exists. It is monic because every two maps into $0$ are equal, so by [L3] it represents a subobject of $A$. Every monomorphism into $A$ factors through $1_A$, and $0\to A$ factors through every monomorphism into $A$, so these classes are respectively the top and bottom elements.

Steps 1.1 and 1.2 prove that the subobjects of $A$ form a bounded lattice.

Consider any lattice in which comparable complements of a fixed element coincide in every interval. Let $X\le Z$ and put $U:=X\vee (Y\wedge Z),V:=(X\vee Y)\wedge Z\mathrm{.}$ Then $U\le V$ by monotonicity.

Fix an interval $[B_1,B_2]$ of subobjects of $A$, and write $Q:=B_2\mathrm{/}{B}_1$ with quotient map $q:B_2\to Q$. If $D$ is any intermediate subobject, let $q_D:B_2\to B_2\mathrm{/}D$ be its quotient map. Because $B_1\le D$, the composite $q_D$ kills $B_1$, so it factors through $q$ as a map ${\bar{q}}_D:Q\to B_2\mathrm{/}D$. Define $\mathrm{\Phi }(D)$ to be the kernel subobject of ${\bar{q}}_D$.

Both $U$ and $V$ are complements of $Y$ in the interval $[Y\wedge Z,X\vee Y]$: one has $U\vee Y=X\vee Y=V\vee Y$, and $Y\wedge Z\le U\le Z$ gives $U\wedge Y=Y\wedge Z$, while $V\le Z$ and $Y\le X\vee Y$ give $V\wedge Y=Y\wedge Z$. By step 1.1, the interval hypothesis forces $U=V$. Therefore the modular-law identity of [L1] holds.

Conversely, if $S\le Q$ with quotient map $t_S:Q\to Q\mathrm{/}S$, define $\mathrm{\Psi }(S)$ to be the kernel subobject of $t_{S}q$ in $B_2$. Then $B_1\le \mathrm{\Psi }(S)\le B_2$. If $D$ lies in the interval, the equality $q_D={\bar{q}}_{D}q$ gives $\mathrm{\Psi }(\mathrm{\Phi }(D))=D$. If $S\le Q$, the morphism $t_{S}q$ is epic, so its image is all of $Q\mathrm{/}S$; by [L3], $B_2\mathrm{/}\mathrm{\Psi }(S)\cong Q\mathrm{/}S\mathrm{.}$ Under this identification, ${\bar{q}}_{\mathrm{\Psi }(S)}$ is a cokernel of $S$, so $\mathrm{\Phi }(\mathrm{\Psi }(S))=S$. Thus $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse bijections between $[B_1,B_2]$ and the subobject lattice of $Q$.

If $D\le E$ in the interval, then $q_E$ factors through $q_D$, so ${\bar{q}}_E$ kills $\mathrm{\Phi }(D)$. Hence $\mathrm{\Phi }(D)\le \mathrm{\Phi }(E)$. The same argument applied to $\mathrm{\Psi }$ shows that $\mathrm{\Phi }$ is an order isomorphism from $[B_1,B_2]$ onto $\mathrm{Sub}(Q)$.

By step 3.1, it is enough to prove the comparable-complements property in $\mathrm{Sub}(Q)$. Let $E\le Q$, and let $U\le V$ be two complements of $E$. Write $t:Q\to Q\mathrm{/}E$ for the quotient map.

Because $U\wedge E=0$, the pullback description [L7] makes the kernel of the restricted map $t{\mathrm{\mid }}_U:U\to Q\mathrm{/}E$ trivial. Because $U\vee E=Q$, the canonical map $[i_U,i_E]:U\oplus E\to Q$ has image $Q$ by [L6], so composing with $t$ shows that $t{\mathrm{\mid }}_U$ is epic. By [L5], $t{\mathrm{\mid }}_U$ is an isomorphism. The same argument shows that $t{\mathrm{\mid }}_V$ is an isomorphism. Let $i:U\to V$ represent the comparison $U\le V$. Then $t{\mathrm{\mid }}_U=(t{\mathrm{\mid }}_V)i$, so $i$ is an isomorphism. Therefore $U$ and $V$ represent the same subobject of $Q$.

Step 5.1 proves that every interval in the subobject lattice has the comparable-complements property, and step 2.1 shows that this property implies the modular law of [L1]. With [L2], this proves that the subobject lattice of $A$ is modular.

Let $q:A\to A\mathrm{/}C$ be the quotient map. Consider the composite $qb:B\to A\mathrm{/}C$. By [L2], a morphism into $B$ is killed by $qb$ exactly when its composite into $A$ factors through $C$, which is exactly the pullback condition defining $B\wedge C$. So $\ker (qb)=B\wedge C$.

By [L3], step 1.1 gives a canonical isomorphism $B\mathrm{/}(B\wedge C)\cong \mathrm{im}(qb)\mathrm{.}$ The map $q$ kills $C$, so its restriction to the join $B\vee C$ factors through the quotient $(B\vee C)\mathrm{/}C$. Conversely, every summand used in the defining map $[b,c]$ lands in $\mathrm{im}(qb)$ after composing with $q$, because the $C$-summand dies. Hence $\mathrm{im}(qb)$ is exactly the image of $B\vee C$ in $A\mathrm{/}C$, namely $(B\vee C)\mathrm{/}C$.

Combining steps 1.1 and 2.1 yields the canonical isomorphism $(B\vee C)\mathrm{/}C\cong B\mathrm{/}(B\wedge C)$.

Assume $f_{\ast }B\le C$. By [L1], the composite $B\rightarrowtail A\stackrel{f}{\to }{A}^{\mathrm{\prime }}$ factors through the subobject $C\rightarrowtail A^{\mathrm{\prime }}$. The pullback defining $f^{\ast }C$ therefore gives a factorization of $B$ through $f^{\ast }C$, so $B\le f^{\ast }C$.

Assume $B\le f^{\ast }C$. Composing with the pullback leg $f^{\ast }C\to C$ shows that the composite $B\to A^{\mathrm{\prime }}$ factors through $C$. By [L3], the image $f_{\ast }B$ is the least subobject of $A^{\mathrm{\prime }}$ with that property, so $f_{\ast }B\le C$.

Steps 1.1 and 1.2 prove the displayed biconditional, which is exactly the Galois-connection condition of [L2].

For any subobject $B\le A$, $B\le f^{\ast }(C_1\wedge C_2)⟺f_{\ast }B\le C_1\wedge C_2⟺f_{\ast }B\le C_1\text{ and }{f}_{\ast }B\le C_2⟺B\le f^{\ast }{C}_1\text{ and }B\le f^{\ast }{C}_2\mathrm{.}$ By [L2], this says that $f^{\ast }(C_1\wedge C_2)$ is the meet of $f^{\ast }{C}_1$ and $f^{\ast }{C}_2$.

For any subobject $C\le A^{\mathrm{\prime }}$, $f_{\ast }(B_1\vee B_2)\le C⟺B_1\vee B_2\le f^{\ast }C⟺B_1\le f^{\ast }C\text{ and }{B}_2\le f^{\ast }C⟺f_{\ast }{B}_1\le C\text{ and }{f}_{\ast }{B}_2\le C\mathrm{.}$ Again [L2] identifies this with the universal property of the join $f_{\ast }{B}_1\vee f_{\ast }{B}_2$.

Steps 1.1 and 1.2 are exactly the two displayed identities.

Pulling back the zero subobject $0\to A^{\mathrm{\prime }}$ along $f$ produces exactly the kernel square of $f$, so by [L1] and [L2] the inverse image $f^{\ast }(0)$ is $\ker (f)$.

The direct image of the identity subobject $1_A$ is, by [L1], the image of the composite $A\stackrel{1_A}{\to }A\stackrel{f}{\to }{A}^{\mathrm{\prime }}$, which is just the image of $f$ in the sense of [L3].

Therefore kernels and images are exactly inverse and direct images along the morphism $f$.

Put $M:=A_1\wedge B_1$, $U:=A_1\wedge B_0$, and $V:=A_0\wedge B_1$. Since $U\le M$, the second isomorphism theorem [L2] applied to $A_0$ and $M$ inside $A_0\vee M$ gives $\frac{A_0\vee M}{A_0\vee U}\cong \frac{M}{M\wedge (A_0\vee U)}\mathrm{.}$ By modularity [L1] inside the interval below $M$, $M\wedge (A_0\vee U)=(M\wedge A_0)\vee U=V\vee U\mathrm{.}$ So the left quotient is canonically isomorphic to $M\mathrm{/}(V\vee U)$.

Similarly, since $V\le M$, the second isomorphism theorem [L2] applied to $M$ and $B_0$ inside $M\vee B_0$ gives $\frac{M\vee B_0}{V\vee B_0}\cong \frac{M}{M\wedge (V\vee B_0)}\mathrm{.}$ Again modularity yields $M\wedge (V\vee B_0)=V\vee (M\wedge B_0)=V\vee U\mathrm{.}$ So the right quotient is also canonically isomorphic to $M\mathrm{/}(V\vee U)$.

The two quotients in steps 1.1 and 1.2 are canonically isomorphic to the same quotient of $M$, hence to each other. The nested-quotient compatibility of [L3] identifies these isomorphisms with the displayed butterfly quotient comparison.

For $0\le i<m$ and $0\le j\le n$, define $A_{i,j}:=A_i\vee (A_{i+1}\wedge B_j)\mathrm{.}$ Then $A_{i,0}=A_i$ and $A_{i,n}=A_{i+1}$, while $A_{i,j}\le A_{i,j+1}$ for every $j$. Concatenating the chains $A_i=A_{i,0}\le A_{i,1}\le \cdots \le A_{i,n}=A_{i+1}$ over $i=0,\dots ,m-1$ gives a refinement of the $A$-chain. Define $B_{j,i}:=B_j\vee (B_{j+1}\wedge A_i)$ symmetrically; concatenating those chains refines the $B$-chain.

For every cell $(i,j)$, apply [L1] to the pairs $A_i\le A_{i+1}$ and $B_j\le B_{j+1}$. It gives a canonical isomorphism $\frac{A_{i,j+1}}{A_{i,j}}\cong \frac{B_{j,i+1}}{B_{j,i}}\mathrm{.}$ So the successive quotients of the two refinements are paired by the same grid.

Some adjacent terms may coincide, producing zero successive quotients. By [F1], deleting those repetitions leaves equivalent refinements, and the quotient pairing from step 2.1 survives on every nonzero factor. Hence the original two chains admit equivalent refinements.

By [L2], the two composition series admit equivalent refinements.

A composition series has no proper refinement. Indeed, if $A_{i-1}<K<A_i$, then the quotient map $A_i\to A_i\mathrm{/}{A}_{i-1}$ carries $K$ to a nonzero proper subobject of the simple object $A_i\mathrm{/}{A}_{i-1}$, contradicting [L1]. So any refinement of a composition series differs from it only by repeated adjacent terms.

Delete repeated adjacent terms from the equivalent refinements of step 1.1. By step 1.2 this recovers the original two composition series, and the quotient pairing survives. Therefore the original series have the same number of factors, and a permutation matches their factors up to isomorphism.

Assume $B$ and $A\mathrm{/}B$ have finite length. Choose composition series $0=B_0<\cdots <B_r=B,0=C_0<\cdots <C_s=A\mathrm{/}B\mathrm{.}$ Let $q:A\to A\mathrm{/}B$ be the quotient map and put $D_j:=q^{-1}(C_j)$. Then $B=D_0<D_1<\cdots <D_s=A,$ and [L3] identifies each quotient $D_j\mathrm{/}{D}_{j-1}$ with $C_j\mathrm{/}{C}_{j-1}$. So $0=B_0<\cdots <B_r=D_0<D_1<\cdots <D_s=A$ is a composition series of $A$. Therefore $A$ has finite length and $\mathrm{\ell }(A)=r+s=\mathrm{\ell }(B)+\mathrm{\ell }(A\mathrm{/}B)$.

Assume now that $A$ has finite length, with composition series $0=A_0<A_1<\cdots <A_n=A\mathrm{.}$ Intersecting with $B$ gives an increasing chain $0=A_0\wedge B\le A_1\wedge B\le \cdots \le A_n\wedge B=B,$ and quotienting by $B$ gives an increasing chain in $A\mathrm{/}B$. By [L3], each successive factor in either chain is a subquotient of a simple factor $A_i\mathrm{/}{A}_{i-1}$, hence is either $0$ or simple. Deleting repeated adjacent terms therefore yields composition series of $B$ and of $A\mathrm{/}B$.

Step 1.1 proves the extension direction and the displayed additive formula. Step 1.2 proves that finite length passes to subobjects and quotients. The number in the formula is independent of the chosen composition series by [L2].

Let $f:X\to Y$ be a morphism between finite-length objects. Since $\ker (f)\le X$ and $\mathrm{im}(f)\le Y$, [L2] makes both $\ker (f)$ and $\mathrm{im}(f)$ finite length. The quotient $Y\mathrm{/}\mathrm{im}(f)$ is then finite length by [L2], so the cokernel of $f$ is finite length as well.

If $X$ and $Y$ have finite length, then the inclusion $X\rightarrowtail X\oplus Y$ has quotient $Y$. Applying [L2] to that inclusion shows that $X\oplus Y$ has finite length. So the finite-length objects are closed under finite biproducts.

Steps 1.1 and 1.2 are exactly the kernel, cokernel, and finite-biproduct closures required by [L3]. Therefore the full subcategory of finite-length objects is an abelian subcategory.

By [L1] and [L3], condition 1 is equivalent to condition 2: the singleton $\{G\}$ is separating exactly when the one-member family $\mathcal{A}(G,-)$ is faithful.

Assume condition 2, and fix an object $A$. By [L2], form the canonical map $e_A:{\coprod }_{u\in \mathcal{A}(G,A)}G\to A$ whose $u$-th coproduct injection is sent to $u$. Let $q:A\to Q$ be its cokernel. If $q\mathrm{\ne }0$, faithfulness of $\mathcal{A}(G,-)$ gives some $u:G\to A$ with $qu\mathrm{\ne }0$. But $u$ is one of the coproduct components of $e_A$, so $qu=0$ because $q{e}_A=0$, a contradiction. Hence $q=0$, and therefore $e_A$ is epic. So condition 2 implies condition 3.

Assume condition 3. If $f,g:X\to Y$ are distinct, then $h:=f-g\mathrm{\ne }0$. Apply condition 3 to $X$: if $hu=0$ for every $u:G\to X$, then $h{e}_X=0$, and since $e_X$ is epic that would force $h=0$. So some $u:G\to X$ satisfies $hu\mathrm{\ne }0$, equivalently $fu\mathrm{\ne }gu$. Thus $G$ separates maps, hence is a generator by [L1]. Therefore condition 3 implies condition 1.

Steps 1.1, 1.2, and 2.1 prove the equivalence of the three descriptions.

Fix an object $A$. For each subobject $m:B\rightarrowtail A$, let $S_m\subseteq \mathcal{A}(G,A)$ be the subset of those maps $u:G\to A$ that factor through $m$. Because $\mathcal{A}$ is locally small, $\mathcal{A}(G,A)$ is a set, so its power set $\mathcal{P}(\mathcal{A}(G,A))$ is a set as well.

For each subset $S\subseteq \mathcal{A}(G,A)$, use AB3 and [L1] to form the canonical map $e_S:{\coprod }_{u\in S}G\to A$ and let $i_S:I_S\rightarrowtail A$ be its image. If $m:B\rightarrowtail A$ is any subobject, then [L1] applied to $B$ gives an epic canonical map ${\coprod }_{v\in \mathcal{A}(G,B)}G\twoheadrightarrow B\mathrm{.}$ Composing with $m$ produces exactly the family of maps in $S_m$, so the image of the resulting composite is $m$. Hence $i_{S_m}$ represents the same subobject as $m$.

The set of monomorphisms $\{i_S:I_S\rightarrowtail A\mid S\subseteq \mathcal{A}(G,A)\}$ therefore contains a representative of every subobject class of $A$. By [L2], $\mathcal{A}$ is well-powered.

By [L1], every object of $\mathcal{A}$ has a set of representative monomorphisms for its subobject classes. By [L3], taking cokernels transfers these to representative epimorphisms for all quotient-object classes. Thus $\mathcal{A}$ is both well-powered and co-well-powered.

By [L5], ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$ is abelian, and cocompleteness of $\mathcal{A}$ becomes completeness of ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$. Local smallness is unchanged. The separating property of $G$ from [L4] becomes the coseparating property in the opposite category.

A supplied well-powering of $\mathcal{A}$ gives a supplied family of representative monomorphisms. Applying cokernels objectwise using [L3] gives a supplied co-well-powering of $\mathcal{A}$, which is a supplied well-powering of ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$. Hence the domain-side hypotheses in branch 1 of [L2] hold for ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$.

Therefore the category supplies exactly the stated dual SAFT data. As [L2] requires, any application must still provide a locally small target and a functor preserving all small limits; those are not consequences of the present category-level hypotheses.

If $B\le C$, then any map $G\to A$ factoring through $B$ also factors through $C$ by composition.

To prove the converse, assume $B\nleqq C$. By [L2], let $p:P\rightarrowtail B$ be the pullback subobject of $b$ and $c$. If $p$ were epic, then [L3] would make it an isomorphism, forcing $b$ to factor through $c$. So $p$ is not epic.

Let $q:B\to Q$ be a cokernel of $p$, which exists by [L5]. Since $p$ is not epic, [L4] implies $q\mathrm{\ne }0$.

The morphisms $q$ and $0_{B,Q}$ are therefore distinct, so [L1] gives some $u:G\to B$ with $qu\mathrm{\ne }0$. If $bu$ factored through $c$, the pullback property in [L2] would force $u$ to factor through $p$, hence $qu=0$, impossible. Thus $bu:G\to A$ factors through $B$ but not through $C$.

Step 1.1 proves the forward implication, and steps 1.2, 2.1, and 3.1 prove the contrapositive of the reverse implication. Therefore $B\le C$ exactly when every morphism $G\to A$ factoring through $B$ also factors through $C$.

Let $(m_i:B_i\rightarrowtail A_i{)}_{i\in I}$ be a small family of monomorphisms, and let $m:{\coprod }_{i\in I}{B}_i\to {\coprod }_{i\in I}{A}_i$ be the induced coproduct map, which exists by the AB3 part of [L1]. Let $k:K\rightarrowtail {\coprod }_i{B}_i$ be its kernel. For each finite subset $F\subseteq I$, let $S_F\le {\coprod }_i{B}_i$ be the finite partial sum of the summands with indices in $F$. Because finite coproducts in an abelian category are biproducts, the restriction of $m$ to $S_F$ is a finite direct sum of monomorphisms and is therefore monic. Hence $K\wedge S_F=0$ for every finite $F$.

The family $(S_F)$ is directed and has join ${\coprod }_i{B}_i$. Applying the AB5 identity [L1] with the fixed subobject $K$ gives $K=\left({\bigvee }_F{S}_F\right)\wedge K={\bigvee }_F(S_F\wedge K)=0.$ So the kernel of $m$ is zero, and therefore $m$ is monic. By [L2], this is exactly AB4.

By [L1], the category $R\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$ has all small coproducts, so it satisfies AB3. For any left $R$-module $M$, the bijection [F1] identifies the canonical coproduct ${\coprod }_{u\in {\mathrm{Hom}}_R(R,M)}R$ with a copy of $R$ for each element of $M$, and the canonical map to $M$ sends the basis vector indexed by $m$ to $m$. It is therefore surjective, hence epic. By [L2], $R$ is a generator.